Titanium(III) chloride, a substance used in catalysts for preparing polyethylene, is made by high-temperature reaction of TiCl₄ vapor with H₂: 2 TiCl₄(g) + H₂(g) → 2 TiCl₃(s) + 2 HCl(g) (a) How many grams of TiCl₄ are needed for complete reaction with 155 L of H₂ at 435 °C and 795 mm Hg pressure?

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Welcome back everyone balloons are at times inflated with ethylene gas. From the reaction of calcium carbide with water as shown below, we have calcium carbide solid reacting with two moles of liquid water to form calcium calcium to hydroxide solid. And our ethylene gas, calculate the mass of calcium carbide required to fill a balloon with ethylene gas to a volume of 23.5 liters and a pressure of 950 millibars at 23 degrees Celsius. Recall that we can utilize our ideal gas equation to solve this problem, which relates our pressure of our gas times are gas times its volume, sorry, equal to our molds of our gas end times. R gas constant R times r temperature in kelvin. So if we know our temperature should be in kelvin, recall that our pressure in this formula should be in units of a T. M. And our volume is going to be in units of leaders recall that our units for our moles of our gas. Our leaders times 80 M, divided by moles times kelvin. So with these units outlined, recognized that were given a pressure As 950 MB. And so we want to convert from millibars in the denominator. Two bars in the numerator, recall that our prefix milli tells us that we have 10 to the negative third power of our base unit. Bar canceling out millibars. We're now going to go from bars to A T. M. Since that's what we use in our ideal gas equation. So recall that we have 1.13 25 bars equivalent to 80 M. And sorry, this is in our denominator, 1.13 25 bars again equivalent to 1 80 M. Canceling out our units of bars were left with a tm for our pressure that we must fill our balloon too. And we're going to find that we have a pressure equal to now 800.9376 A. T. M's to fill balloon. Now, with this pressure solved, we need our next unit from our prompt being our temperature in C were given 23°C. so we need this in kelvin. So we're going to add to 73.15 and we'll get our kelvin temperature of 96.15 kelvin. Now, with our proper temperature and pressure units, we want to find our moles of our gas and so we need to manipulate our ideal gas equation so that we can isolate for moles of gas and so we would have our moles of our only gas in the prompt being ethylene gas C two H two equal to our pressure times, volume divided by r gas constant R times our temperature in kelvin. And so solving for and we just want to plug everything in. So we would have our pressure which we just found as 0.9376 A tms of pressure Times are volume, which in the prompt is given as 23.5 liters required of ethylene gas to fill the balloon. This is already in the proper units. And then we will divide by r gas constant R. Which we should recall is the value 0.8 to 06 units of leaders times A. T. M. Divided by moles times kelvin. And then multiplying by this. We have our temperature which we converted to 2 96. Kelvin. And so canceling out our units. Notice we can get rid of A. T. M. We can cancel out kelvin and we can cancel out leaders. And we're left with moles as our final unit, which is what we want. And we would find that our moles of ethylene gas required to fill the balloon is equal 2.9066 moles of C two H two ethylene. Now with our moles of ethylene gas, we can figure out our mass of calcium carbide required to fill the balloon by using our molar ratio from our balanced equation, in which our ratio between calcium carbide to ethylene gas we can see is a 1 to 1 molar ratio from our balanced equation. And so utilizing our moles of ethylene, we want to find again our mass of our calcium calcium carbide. And so the master cried to fill the balloon beginning with moles of ethylene. We have 0.9066 moles of C. Two H two which we just solved above. And multiplying now by our molar ratio to go from moles of Kathleen and the denominator, two moles of calcium carbide in the numerator. We said we have a 1 to 1 molar ratio. So now we can cancel out moles of ethylene. We're left with moles of calcium carbide, but we need our final unit to actually be mass. And so we're not going to stop here. We're going to continue and multiply by our next conversion factor to go from moles of calcium carbide in the denominator, two g of calcium carbide in the numerator and utilizing our molar mass of calcium carbide. We would find in our periodic table. We have a mass of 64.0 98 g of calcium carbide equivalent to one mole. And so now canceling out moles of calcium carbide. We're left with grams of calcium carbide as our final unit. And this will yield a mass equal to 58.1 g of calcium carbide required to fill the balloon with water. And so our final answer is going to be this mass here highlighted in yellow of calcium carbide, which will correspond to choice be in the multiple choice as the correct answer. So I hope that everything I went through is clear and let us know if you have any questions

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Key Concepts

Ideal Gas Law

Stoichiometry

Molar Mass